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Please Help with Angular Momentum/Angular Speed

  1. Jun 23, 2004 #1
    A person exerts a tangential force of 36.1 N on the rim of a disk-shaped merry-go-round of radius 2.74 m and mass 167 kg. If the merry-go-round starts at rest, what is its angular speed after the person has rotated it through an angle of 60.0°?

    => L=Iw => w = L/I where I=1/2mr^2
    I just don't know where to plug in the tangential force and the angle.
    Please help!
     
  2. jcsd
  3. Jun 24, 2004 #2
    Let F be the tangential force, R the radius of the merry-go-round and M its mass. The torque is [itex]\tau = FR = I\alpha[/itex]. You know F, R and I so you can solve for [itex]\alpha[/itex] and then use that to find the angular velocity. Pretty straight forward. Remember, physics is not about plugging numbers and spitting answers.
     
  4. Jun 24, 2004 #3
    Thank you so very much for your prompt reply.

    theta =1 rad = 57.3degree, 60degree=1.05 rad

    I calculated
    a)alpha = T/I = 98.9/627 = 0.158 rad/s
    b)theta = 1/2 (alpha) t^2 => t = sqrt(2 * 1.05 / 0.158) = 3.65s
    c)w = (alpha) t = 0.158 rad/s^2 * 3.65 s = 0.577 rad/s

    Please let me know if my figures are correct. Again, thank you.
     
  5. Jun 24, 2004 #4
    Instead of finding t, you could just use [itex]\omega^2 = \omega_0^2 + 2\alpha\theta[/itex]. You can check your answer doing it this way.
     
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